# Difference between revisions of "Proofs"

Let $ax^2+bx+c=0$. Then $$x^2+\frac{b}{a}x+\frac{c}{a}=0$$ Completing the square, we get $$\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}$$ Simplifying, we see $$\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

## Pythagorean Theorem

Since area of green square is $a^2$

Since are of blue square is $b^2$

Since red square is $c^2$

We have the following relationship

Based on this, we get we get $\boxed {a^2+b^2=c^2}$ (here we get the Pythagorean Theorem)